Question: Evaluate the iterated integral. $ \int_{-2}^{1} \left( \int_0^3 2x^2 + y^2 - 8xy^3 \, dx \right) dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $26$ (Choice B) B $44$ (Choice C) C $142$ (Choice D) D $198$
Answer: Evaluate the inner integral: $\begin{aligned} \int_{-2}^{1} \left( \int_0^3 2x^2 + y^2 - 8xy^3 \, dx \right) dy &= \int_{-2}^1 \left[ \dfrac{2}{3}x^3 + y^2x - 4x^2y^3 \right]_0^3 dy \\ \\ &= \int_{-2}^1 18 + 3y^2 - 36y^3 \, dy \end{aligned}$ Evaluate the outer integral: $\begin{aligned} \int_{-2}^1 18 + 3y^2 - 36y^3 \, dy &= 18y + y^3 - 9y^4 \bigg|_{-2}^1 \\ \\ &= (18 + 1 - 9) - (-36 - 8 - 144) \\ \\ &= 198 \end{aligned}$ The answer: $ \int_{-2}^{1} \left( \int_0^3 2x^2 + y^2 - 8xy^3 \, dx \right) dy = 198$